Question

The expression (a-b)³+(b-c)³+(c-a)³ can be factorized as:
A. (a-b)(b-c)(c-a)
B. 3(a-b)(b-c)(c-a)
C. -3(a-b)(b-c)(c-a)
D. (a+b+c)(a²+b²+c²-ab-bc-ca)

Answer 1

The algebraic expression is given as : (a - b)³ + (b - c)³ + (c - a)³

Let (a - b) = a , (b - c) = b , (c - a) = c. Then  

a + b + c =  a - b + b - c + c - a = 0

We know, a³ + b³ + c³− 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

∴ a³ + b³ + c³ =  3abc

⇒ (a - b)³ + (b - c)³ + (c - a)³ = 3(a - b) (b - c)(c - a)

Hence, the expression (a - b)³ + (b - c)³ + (c - a)³ can be factorized as 3(a - b) (b - c)(c - a).

Among the given options option (B) 3(a - b)(b - c)(c - a) is correct.

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Answer 2

Answer:

Step-by-step explanation:

Let (a - b) = a , (b - c) = b , (c - a) = c. Then  

a + b + c =  a - b + b - c + c - a = 0

We know, a³ + b³ + c³− 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

∴ a³ + b³ + c³ =  3abc

⇒ (a - b)³ + (b - c)³ + (c - a)³ = 3(a - b) (b - c)(c - a)